Minimum Wind Velocity to Lift a Flat Insulation Panel: 04-BS-7, May 2016, Q4

Question 4: Minimum Wind Velocity to Lift a Flat Insulation Panel (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Panel size	2.438 m × 1.219 m × 25 mm thick
Panel density ρ_p	100 kg/m³
Underside condition	stagnation (grass reduces velocity to zero)
Ambient air density ρ_air (assumed 15°C)	1.21 kg/m³

Find. The minimum wind velocity V (km/hr) that lifts the panel.

Check: the paper gives air density at both 20°C and 15°C; 15°C (1.21 kg/m³) is used here as the ambient reference — using 20°C instead changes the answer by under 1%.

Approach. Apply Bernoulli along the streamline that reaches stagnation under the panel (pressure there = P_atm + full dynamic pressure) and compare with the pressure just above the panel, where the wind still moves at essentially the free-stream velocity V (pressure there ≈ P_atm); the panel lifts when this pressure difference times its area equals its weight.

Pressure difference across the panel. Below (stagnation): $p_{below}=p_{atm}+\tfrac12\rho_{air}V^2$. Above (undisturbed free stream at the same elevation): $p_{above}=p_{atm}$. So the net uplift pressure is $$\Delta p = p_{below}-p_{above} = \tfrac12\rho_{air}V^2$$
Set uplift force equal to panel weight. Panel weight per unit plan area is $\rho_p g t$ (t = thickness), so at lift-off: $$\tfrac12\rho_{air}V^2 = \rho_p g t \quad\Rightarrow\quad V=\sqrt{\frac{2\rho_p g t}{\rho_{air}}}$$
Substitute numbers. $$V = \sqrt{\frac{2(100)(9.81)(0.025)}{1.21}} = \sqrt{40.54} = 6.37\ \text{m/s}$$ $$V = 6.37 \times 3.6 = \boxed{22.9\ \text{km/hr}}$$

Quantity	Result
Minimum lift-off wind velocity V	6.37 m/s = 22.9 km/hr

Topic: Bernoulli-driven lift on a thin, ground-hugging plate (a simple "roof uplift" model).

Key relations: stagnation pressure p₀ = p + ½ρV²; lift-off when ½ρ_airV² = ρ_panelgt.

Why this works. The grass friction kills the flow immediately under the panel, converting all of its dynamic pressure into static pressure there (stagnation), while the smooth wind above the panel keeps most of its speed and stays near atmospheric — the resulting pressure imbalance is exactly the same mechanism that generates lift under a roof overhang or a loose sheet of plywood in a storm.

Common pitfall. Using the panel's plan area or overall mass explicitly is unnecessary — both cancel, leaving lift-off velocity dependent only on thickness and the panel/air density ratio, not on the panel's length or width.

Source: White, Fluid Mechanics, Ch. 3 (Bernoulli equation, stagnation pressure).

Question 4 of 13: Minimum Wind Velocity to Lift a Flat Insulation Panel

Question 4: Minimum Wind Velocity to Lift a Flat Insulation Panel (5 marks)